Analytic functions (part 1)

I am obligated to mention that I made a mistake when talking about holomorphic functions. To check that {f} is holomorphic, you must check that {f} is continuous and the partials {u_x}, etc exist and satisfy the Cauchy-Riemann equations. I forgot to mention that {f} needs to be continuous. The proof relies on the continuity, and I even used this fact in the examples.

Now let’s talk about analytic functions and power series. Let {U} be an open set in {{\mathbb C}}, and let {f:U\rightarrow{\mathbb C}}. Then we say {f} is analytic if locally, it is given by a power series. This is to say, it doesn’t have to have a unique power series representation on all of {U}, but if you pick a point {z_0\in U}, there’s a neighborhood around {z_0} for which {f} is given by a power series centered at {z_0}. By power series centered at {z_0}, I mean that, in this small neighborhood of {z_0},

\displaystyle f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots,

for {a_n\in{\mathbb C}}. Don’t forget that even though we can pretend, we really don’t add up infinitely many terms. What we have really done is defined the finite sum {f_n(z)=a_0+a_1(z-z_0)+\cdots+a_n(z-z_0)^n}, and let

\displaystyle f(z)=\displaystyle\lim_{n\rightarrow\infty}f_n(z).

The distinction is important.

But these power series don’t always converge. In the case of real power series, recall that {1/(1-x)=1+x+x^2+x^3+\cdots}, but this only makes sense where the right-hand side converges (between {1} and {-1}). A similar thing happens over {{\mathbb C}}.

For a power series like the one given above, define

\displaystyle R=\left(\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.

The astute observer will recognize that this could be {1/0}, or {1/\infty}. In such a case, we just say that {R=\infty} or {R=0} respectively. We call {R} the radius of convergence for reasons that the next theorem should make clear. For example, {R=\infty} will mean the power series converges everywhere. There are a few points to note here. If the power series is finite, then for {n} large enough, {a_n=0}, meaning {R=\infty}. This makes sense, because polynomials are well defined on all of {{\mathbb C}}. Also, recognize that this definition is perfectly valid for real power series. In particular, for {1+x+x^2+\cdots}, the radius of convergence is 1.

Lemma 1 (Root Test) Let {f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots} be a power series, and define

\displaystyle R=\left(\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.

Then if {|z-z_0|<R}, {f(z)} converges. If {|z-z_0|>R}, {f(z)} diverges. We make no claims about the case {|z-z_0|=R}.

Proof: We need to treat the cases {R=0} and {R=\infty} separately. I won’t bother doing them at all, because they’re really much easier than the case where {R\in(0,\infty)}. Suppose first that {r=|z-z_0|>R}. Then {\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=1/R>1/r}, meaning there are infinitely many {n} for which {\sqrt[n]{|a_n|}>1/r}. That is, there are infinitely many {n} for which

\displaystyle |a_n|\cdot|z-z_0|^n=|a_n|r^n>1.

Thus, the terms in the sum {\sum_{n=0}^\infty a_n(z-z_0)^n} do not converge to zero, so there is no way the sum could converge.

On the other hand, if {r=|z-z_0|<R}, then {1/r>\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}}. From the definition of {\limsup}, this means there is some point after which all entries are less than {1/r}. More technically, there is some number {N} such that for each {n>N},

\displaystyle \sqrt[n]{|a_n|}<1/r.

We can even do a bit better than that. Just being less than {1/r}, they could get arbitrarily close to {1/r}. Of course, if that were the case, then {1/r} would be the {\limsup}, which it’s not (it’s greater than the {\limsup}). So in fact there is some real number {x<1/r} for for which, whenever {n>N},

\displaystyle \sqrt[n]{|a_n|}<x.

Now, it follows that {|a_n|r^n<x^nr^n}. Since {x\cdot r<1}, we have shown that the terms in the power series (after a certain point) grow slower in magnitude than a geometric series of radius less than one. Thus, {f(z)} must converge.


This was a proof of the root test, and a first step towards what we’ll prove next time: That analytic functions are holomorphic.


6 Responses to Analytic functions (part 1)

  1. Pingback: Analytic functions (part 2) « Andy Soffer

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